OFFSET
1,2
COMMENTS
If the canonical factorization of n into prime powers is the product of p^e(p) then sigma_k(n) = Product_p ((p^((e(p)+1)*k))-1)/(p^k-1).
Sum_{d|n} 1/d^k is equal to sigma_k(n)/n^k. So sequences A017665-A017712 also give the numerators and denominators of sigma_k(n)/n^k for k = 1..24. The power sums sigma_k(n) are in sequences A000203 (k=1), A001157-A001160 (k=2,3,4,5), A013954-A013972 for k = 6,7,...,24. - Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 05 2001
LINKS
Seiichi Manyama, Table of n, a(n) for n = 1..10000 (first 1000 terms from Harvey P. Dale)
FORMULA
G.f.: Sum_{k>=1} k^23*x^k/(1-x^k). - Benoit Cloitre, Apr 21 2003
From Amiram Eldar, Oct 29 2023: (Start)
Multiplicative with a(p^e) = (p^(23*e+23)-1)/(p^23-1).
Dirichlet g.f.: zeta(s)*zeta(s-23).
Sum_{k=1..n} a(k) = zeta(24) * n^24 / 24 + O(n^25). (End)
MATHEMATICA
DivisorSigma[23, Range[15]] (* Harvey P. Dale, May 02 2016 *)
PROG
(Sage) [sigma(n, 23)for n in range(1, 12)] # Zerinvary Lajos, Jun 04 2009
(PARI) vector(30, n, sigma(n, 23)) \\ G. C. Greubel, Nov 03 2018
(Magma) [DivisorSigma(23, n): n in [1..30]]; // G. C. Greubel, Nov 03 2018
CROSSREFS
KEYWORD
nonn,easy,mult
AUTHOR
STATUS
approved